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Transcript of Chapter 6 Series and Parallel Circuits 1 of 81 MECH1100 1 of 81 Chapter 6 Series and Parallel...
Chapter 6 Series and Parallel Circuits
1 of 81
MECH1100
1 of 81
Chapter 6 Series and Parallel Circuits MECH1100
TopicsIdentifying Series- Parallel Relationships
Analysis of Series-Parallel Resistive Circuits
Voltage Dividers with Resistive Loads
Loading Effect of a Voltmeter
Wheatstone Bridge
Thevenin’s Theorem
Maximum Power Transfer
Superposition
Chapter 6 Series and Parallel Circuits
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Most practical circuits have combinations of series and parallel components.
Identifying series-parallel relationships
Components that are connected in series will share a common path.
Components that are connected in parallel will be connected across the same two nodes.
1 2
From Chapters 4 and 5
Chapter 6 Series and Parallel Circuits
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You can frequently simplify analysis by combining series and parallel components.
Combination circuits
Solve by forming the simplest equivalent circuit possible.
1. has characteristics that are electrically the same as another circuit
2. is generally simpler.
Circuits containing both series and parallel circuits are called COMBINATION circuits
An equivalent circuit is one that :
Chapter 6 Series and Parallel Circuits
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Equivalent circuits
R 1
R 2R 2
1
1 .0 k
1 .0 k
is equivalent to R 11
2 .0 k
There are no electrical measurements that can distinguish between the boxes.
Chapter 6 Series and Parallel Circuits
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Another example:
There are no electrical measurements that can distinguish between the boxes.
R R1 2
1 .0 k 1 .0 kR
1 ,2
5 0 0
is equivalent to
Equivalent circuits
Chapter 6 Series and Parallel Circuits
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R
R1
R 2R 2
31 .0 k
4 .7 k2 .7 k
is equivalent to
R R1 ,2 3
4 .7 k3 .7 k
R 1 ,2 ,3
2 .0 7 k
is equivalent to
There are no electrical measurements that can distinguish between the three boxes.
Equivalent circuits
Chapter 6 Series and Parallel Circuits
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What Do We Know1. For Series Circuits:
a. Current at all points is the IT = IR1 = IR2
b. Voltage across each resistorVT = V1 + V2
2. For Parallel Circuita. Current
IT = IR1 + IR2
b. Voltage at all nodes is the VT = V1 = V2
same
same
drops
divides across each resistor in the branch
Chapter 6 Series and Parallel Circuits
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Seven Step Process for Solving a Combination Circuit
1.Simplify the circuit to a series circuit by finding the effective equivalent resistance (REQ) of each parallel section in the circuit. Redraw the circuit each time it is simplified.
2.Calculate the total resistance (RT) of the circuit by adding all REQ’s to the other series resistances.
3.Calculate the total current (IT) using RT in Ohm’s law.
4.Calculate the voltage drop across any series resistances or REQ’s using Ohm’s law.
5.Calculate the branch currents in all parallel sections of the circuit using the voltage drop across REQ and Ohm’s law.
6.Use the branch currents and resistance values to calculate the voltage of the parallel resistances.
7.Make a summary of the voltage drops and currents for each resistance to make sure they total correctly.
Chapter 6 Series and Parallel Circuits
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A simple series-parallel resistive circuit.
Chapter 6 Series and Parallel Circuits
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R4 is added to the circuit in series with R1.
Chapter 6 Series and Parallel Circuits
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R5 is added to the circuit in series with R2.
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R6 is added to the circuit in parallel with the series combination of R1 and R4.
Chapter 6 Series and Parallel Circuits
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Sketch the circuit and write the equation for
nodes A to B
)( 321 RRRRT
Chapter 6 Series and Parallel Circuits
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)( 213 RRRRT
Sketch the circuit and write the equation for
nodes A to C
Chapter 6 Series and Parallel Circuits
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)3( 12 RRRRT
Sketch the circuit and write the equation for
nodes B to C
Chapter 6 Series and Parallel Circuits
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Draw the schematic for this circuit board and give the
equation for the relationship of the resistor
branches
Chapter 6 Series and Parallel Circuits
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7412365)( )())(( RRRRRRRR ABEQ
Chapter 6 Series and Parallel Circuits
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Reduce the following circuit to REQ
Chapter 6 Series and Parallel Circuits
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)( 43 RRREQ
6 Ω
Chapter 6 Series and Parallel Circuits
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643 )( RRRREQ
10 Ω6 4
Chapter 6 Series and Parallel Circuits
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5643 ))(( RRRRREQ 5 Ω10 10
Chapter 6 Series and Parallel Circuits
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215643 )))((( RRRRRRREQ 50 ΩIT = 2
amps
IT
5 15 30
Chapter 6 Series and Parallel Circuits
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Review of voltage relationships
688ABR
647BCR
1334)( ACEQR
15.510*1334
687)( ABV
85.410*1334
647)( BCV
107.49
1334TI mA
Chapter 6 Series and Parallel Circuits
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I
+
+
26.5 mA
I
+18.5 mA
I
+8.0 mA
R5100
R3330
R2470
R1270
VS5.0 V
R4
100
R6
100
A- -
-
Kirchoff’s current law
What are the readings for node A?
Chapter 6 Series and Parallel Circuits
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Loaded voltage divider
1. A voltage-divider with a resistive load forms a combination (parallel) circuit.
The voltage-divider equation was developed for a series circuit. Recall that the output voltage is given by
A
22 S
T
RV V
R
R1
R2 R3
+
2. The voltage divider is said to be LOADED.
3. The loading reduces the total resistance from node A to ground.
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Loaded voltage divider
A
What is the voltage across R3?
Form an equivalent series circuit by combining R2 and R3; then apply the voltage-divider formula to the equivalent circuit:
2,33 2,3 S
1 2,3
387 15 V
330 387 R
RV V V
R R
+15 V R1
R2 R3
330
470 2.2 k
VS =
2,3 2 3 470 2.2 k = 387 R R R
8.10 V
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Stiff voltage divider
• A stiff voltage-divider is one in which the loaded voltage is nearly the same as the no-load voltage.
• To accomplish this, the load current must be small compared to the bleeder current (or RL is large compared to the divider resistors).
R1
R2 RL
VS
21 & RRRL
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Stiff voltage divider
R1
R2 RL
VS
Given: R1 = R2 = 1.0 kW, 1. What value of RL will
make the divider a stiff voltage divider?
2. What fraction of the unloaded voltage is the loaded voltage?
Rule of Thumb: RL = R2*10; Thus: RL should be 10 kW or greater.
2 LRL S S S
1 2 L
|| 0.91 k 0.476
|| 1.0 k 0.91 k
R RV V V V
R R R
This is 95% of the unloaded voltage.
SSSloadnoR VVkk
kV
RR
RV 5.
11
1
21
2)(2
-
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Wheatstone bridge
Voltage Divider 1
Voltage Divider 2
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Wheatstone Bridge
V1=V2
I1=I3
V3=V4
I2=I4
4
231
4
2
3
1
R
RRR
R
R
R
RL R
L R
U U
L L
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Wheatstone Bridge
• The Wheatstone bridge consists of:1. a dc voltage source and 2. four resistive arms
forming two voltage dividers.
• The output is taken between the dividers.
• Frequently, one of the bridge resistors is adjustable. (R2)
-
+
R 1R 3
R 4R 2
V SO u tpu t
.
+4.04 V
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Balanced Wheatstone Bridge
When the bridge is balanced, the output voltage is
-
+
R 1R 3
R 4R 2
V SO u tpu t
zero
The products of resistances in the opposite diagonal arms are equal.
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Wheatstone bridge
-
+
R 1R 3
R 4R 2
V SO u tpu t
What is the value of R2 if the bridge is balanced?
470 W 330 W
270 W
12 V
385 W
4
321
R
RRR
Chapter 6 Series and Parallel Circuits
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Measuring a physical parameter using a transducer.
Unbalanced Wheatstone Bridge
• Unbalance occurs when VOUT ≠ 0
• Used to measure1. Mechanical Strain2. Temperature3. Pressure
• VOUT is converted to a digital output indicating the value of the reading.
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ST
XX V
R
RV
VOUT = VA-VB = 0
Chapter 6 Series and Parallel Circuits
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10
10
10
10
10
Using the Thermistor chart, what is VOUT when the temperature is 50o C
VA=8.8v VB=6.0v VA-B=2.8v
3
3A S
RV V
R Rtherm
SB VRR
RV
42
4
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Example of a load cell
Chapter 6 Series and Parallel Circuits
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Load Cell or Strain Gauge
Chapter 6 Series and Parallel Circuits
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Typical load cell (strain gauge) schematic
Chapter 6 Series and Parallel Circuits
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Typical load cell (strain gauge) schematic
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Typical load cell (strain gauge) schematic
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Troubleshooting
The effective troubleshooter must think logically about circuit operation.
Understand normal circuit operation and find out the symptoms of the failure.
Decide on a logical set of steps to find the fault.
Following the steps in the plan, make measurements to isolate the problem. Modify the plan if necessary.
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Troubleshooting
The output of the voltage-divider is 6.0 V. Describe how you would use analysis and planning in finding the fault.
From an earlier calculation, V3 should equal 8.10 V. A low voltage is most likely caused by a low source voltage or incorrect resistors (possibly R1 and R2 reversed). If the circuit is new, incorrect components are possible.
Decide on a logical set of steps to locate the fault. You could decide to 1) check the source voltage, 2) disconnect the load and check the output voltage, and if it is correct, 3) check the load resistance. If R3 is correct, check other resistors.
A+15 V R1
R2 R3
330
470 2.2 k
VS =
Chapter 6 Series and Parallel Circuits
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Loading
Load current
Bleeder current
Wheatstone bridge
The current left after the load current is subtracted from the total current into the circuit.
The output current supplied to a load.
The effect on a circuit when an element that draws current from the circuit is connected across the output terminals.
A 4-legged type of bridge circuit with which an unknown resistance can be accurately measured using the balanced state. Deviations in resistance can be measured using the unbalanced state.
Key Terms
Chapter 6 Series and Parallel Circuits
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Thevenin’s theorem
Maximum power transfer
Superposition
The condition, when the load resistance equals the source resistance, under which maximum power is transferred to the load.
A method for analyzing circuits with two or more sources by examining the effects of each source by itself and then combining the effects.
A circuit theorem that provides for reducing any two-terminal resistive circuit to a single equivalent voltage source in series with an equivalent resistance.
Key Terms
Chapter 6 Series and Parallel Circuits
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1. Two circuits that are equivalent have the same
a. number of components
b. response to an electrical stimulus
c. internal power dissipation
d. all of the above
Quiz
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2. If a series equivalent circuit is drawn for a complex circuit, the equivalent circuit can be analyzed with
a. the voltage divider theorem
b. Kirchhoff’s voltage law
c. both of the above
d. none of the above
Quiz
Chapter 6 Series and Parallel Circuits
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3. For the circuit shown,
a. R1 is in series with R2∥R3b. R1 is in parallel with R2
c. R2 is in series with R3
d. R2 is in parallel with R1 + R3
-
+
R 1
R 3
R 2
V S
Quiz
Chapter 6 Series and Parallel Circuits
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4. For the circuit shown,
a. R1 is in series with R2
b. R4 is in parallel with R1
c. R2 is in parallel with R3
d. none of the above
-
+R 1
R 4
R 3
R 2
V S
Quiz
Chapter 6 Series and Parallel Circuits
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6. For the circuit shown, Kirchhoff's voltage law
a. applies only to the outside loop
b. applies only to the A junction.
c. can be applied to any closed path.
d. does not apply. R1
R3
470
270
R2
330
VS +10 V A
Quiz
Chapter 6 Series and Parallel Circuits
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7. The effect of changing a measured quantity due to connecting an instrument to a circuit is called
a. loading
b. clipping
c. distortion
d. loss of precision
Quiz
Chapter 6 Series and Parallel Circuits
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8. An unbalanced Wheatstone bridge has the voltages shown. The voltage across R4 is
a. 4.0 V
b. 5.0 V
c. 6.0 V
d. 7.0 V
-
+
R L
R 1R 3
R 4R 2
V S
12 V
1 .0 V
7 .0 V+ -
VR2 = 5V
VR2 – VRL = 4V
VR3 = 8V
Quiz
Chapter 6 Series and Parallel Circuits
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9. Assume R2 is adjusted until the Wheatstone bridge is balanced. At this point, the voltage across R4 is measured and found to be 5.0 V. The voltage across R1 will be
a. 4.0 V
b. 5.0 V
c. 6.0 V
d. 7.0 V
-
+
RL
R1R3
R4R2
VS
12 V
5.0 V
+ -
Quiz
Chapter 6 Series and Parallel Circuits
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10. Maximum power is transferred from a fixed source when
a. the load resistor is ½ the source resistance
b. the load resistor is equal to the source resistance
c. the load resistor is twice the source resistance
d. none of the above
Quiz